HA + HB + HC $\geq$ \[\sqrt[3]{{\frac{{4abc}}{{\sin A + \sin B + \sin C}}}} + \frac{{\sqrt[3]{{2{a^2}{b^2}{c^2}(\sin 2A + \sin 2B + \sin 2C)}}}}{{a + b + c}}\]
Edited by Hieutran2000, 08-11-2015 - 18:09.
Cho tam giác ABC, trực tâm H, Kí hiệu S là diện tích
Started By Hieutran2000, 07-11-2015 - 22:17
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Posted 07-11-2015 - 22:17
Hieutran2000
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Hạ sĩ
Cho tam giác ABC, trực tâm H. Chứng minh:
HA + HB + HC $\geq$ \[\sqrt[3]{{\frac{{4abc}}{{\sin A + \sin B + \sin C}}}} + \frac{{\sqrt[3]{{2{a^2}{b^2}{c^2}(\sin 2A + \sin 2B + \sin 2C)}}}}{{a + b + c}}\]
HA + HB + HC $\geq$ \[\sqrt[3]{{\frac{{4abc}}{{\sin A + \sin B + \sin C}}}} + \frac{{\sqrt[3]{{2{a^2}{b^2}{c^2}(\sin 2A + \sin 2B + \sin 2C)}}}}{{a + b + c}}\]
$\sum =\prod$
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